Help with Lagrange's theorem.

Lagrange's Theorem states that: 'If H is a sub-group of a finite number G, then the order of H is a factor of G'.
It follows that 'The period of an element is a factor of the order of the group.'
Use the two theorems above and you knowlegde of groups to justify the proposition that only 2 groups of the order 6 exists, one Abelian and one non-Ablien. (please DON'T use cosets, and do use identities and subgroups or periods)

Lagrange's Theorem states that: 'If H is a sub-group of a finite number G, then the order of H is a factor of G'.
It follows that 'The period of an element is a factor of the order of the group.'
Use the two theorems above and you knowlegde of groups to justify the proposition that only 2 groups of the order 6 exists, one Abelian and one non-Ablien. (please DON'T use cosets, and do use identities and subgroups or periods)

Provision of the supporting arguments in the form of proof
Thanks

This is not the answer but opinions :

I read an essay from Arthur Cayley , the only three groups of order six are :

and

But there are two possible groups :

(commute)

however , the first possible group is isomorphic to the group first mentioned with because

Case one: There is an element of order / period 6. Then, generates the group. Therefore, the group is cyclic and is isomorphic to .

Case two: There are no elements of order / period 6. This means that the only possible orders / periods are 1, 2, 3, by Lagrange's Theorem. Suppose is an element of order 3. It forms a subgroup consisting of . Take note that has order 3 as well. This means that order / period 3 elements come in pairs. Since there is one identity element, we only have 5 elements to "play with". Combining these two facts means that there can be only 2 or 4 order / period 3 elements. We will have trouble if there are 4 of them. (Try to prove this on your own!) This leaves 3 elements, all of which have to be order 2, by exhaustion.

In summary, we are forced to have 1 identity, 3 order / period 2, and 2 order / period 3 elements. Such a group is isomorphic to , the group of permutations on three letters.